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Answer» Area Under the Curve mathematically means that we have to find the definite integral with the given limits.
We asked to find the area under the curve  , the X-Axis, and between  and 
In other words, we have to find the Integral of  with limits as to
[A graph is attached for visual representation]
Let the Area be 
Then we have:
![\displaystyle A = \int\limits_0^2 5e^x \, dx \\ \\ \\ \implies A=5\left[\, e^x\, \right]_0^2 \\ \\ \\ \implies A=5[e^2-e^0]\\ \\ \\ \implies \boxed{\bold{A=5(e^2-1) \, \, sq. \, units}}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20A%20%3D%20%5Cint%5Climits_0%5E2%205e%5Ex%20%5C%2C%20dx%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20A%3D5%5Cleft%5B%5C%2C%20e%5Ex%5C%2C%20%5Cright%5D_0%5E2%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20A%3D5%5Be%5E2-e%5E0%5D%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cboxed%7B%5Cbold%7BA%3D5%28e%5E2-1%29%20%5C%2C%20%5C%2C%20sq.%20%5C%2C%20units%7D%7D%20 "\displaystyle A = \int\limits_0^2 5e^x \, dx \\ \\ \\ \implies A=5\left[\, e^x\, \right]_0^2 \\ \\ \\ \implies A=5[e^2-e^0]\\ \\ \\ \implies \boxed{\bold{A=5(e^2-1) \, \, sq. \, units}}")
Thus, The Answer is Option (C).
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In the second QUESTION, we are given that a car starts from rest, and moves with constant ACCELERATION.
Let us suppose the constant acceleration  .
Let distance covered in n seconds be denoted by 
Then, we have:
Now, We need the RATIO of distance travelled in  second (say  ) to the distance travelled in  seconds ()
To get the distance travelled in second, we will use this logic:
Distance travelled in the nth second is EQUAL to the distance travelled in (n-1) seconds subtracted from the distance travelled in n seconds.
That is,
For EXAMPLE, suppose we want to find the distance travelled in the 4th second. Then we first find the distance travelled in complete 4 seconds, and from it, we subtract the distance travelled in 3 seconds.
Now, we can get the answer.
![\displaystyle Ans = \frac{s_{n^{th}}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{s_n-s_{n-1}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{\frac{an^2}{2}-\frac{a(n-1)^2}{2}}{\frac{an^2}{2}}\\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{n^2-(n-1)^2}{n^2} \\ \\ \\ \left[\text{Using the Identity }a^2-b^2=(a+b)(a-b)\right] \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(n+n-1)(n-n+1)}{n^2} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(2n-1)(1)}{n^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Ans%20%3D%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%3D%20%5Cfrac%7Bs_n-s_%7Bn-1%7D%7D%7Bs_n%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%3D%20%5Cfrac%7B%5Cfrac%7Ban%5E2%7D%7B2%7D-%5Cfrac%7Ba%28n-1%29%5E2%7D%7B2%7D%7D%7B%5Cfrac%7Ban%5E2%7D%7B2%7D%7D%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%3D%20%5Cfrac%7Bn%5E2-%28n-1%29%5E2%7D%7Bn%5E2%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cleft%5B%5Ctext%7BUsing%20the%20Identity%20%7Da%5E2-b%5E2%3D%28a%2Bb%29%28a-b%29%5Cright%5D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%3D%20%5Cfrac%7B%28n%2Bn-1%29%28n-n%2B1%29%7D%7Bn%5E2%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cfrac%7Bs_%7Bn%5E%7Bth%7D%7D%7D%7Bs_n%7D%20%3D%20%5Cfrac%7B%282n-1%29%281%29%7D%7Bn%5E2%7D "\displaystyle Ans = \frac{s_{n^{th}}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{s_n-s_{n-1}}{s_n} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{\frac{an^2}{2}-\frac{a(n-1)^2}{2}}{\frac{an^2}{2}}\\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{n^2-(n-1)^2}{n^2} \\ \\ \\ \left[\text{Using the Identity }a^2-b^2=(a+b)(a-b)\right] \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(n+n-1)(n-n+1)}{n^2} \\ \\ \\ \implies \frac{s_{n^{th}}}{s_n} = \frac{(2n-1)(1)}{n^2}")
Thus, The answer is Option (C)
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